Chapter 1: Introduction and Basic Concepts
Çengel/Boles - Thermodynamics: An Engineering Approach, 8 th Edition
(Numerical values for solutions can be obtained by copying the EES solutions given and
pasting them on a blank EES screen, and pressing the Solve command. Similar
problems and their solutions can be obtained easily by modifying numerical values.)
Chap1-1 Pressure Difference in Water (Submarine)
Consider a submarine cruising 30 m below the free surface of seawater whose density is
1025 kg/m3. The increase in the pressure exerted on the submarine when it dives to a
depth of 110 m below the free surface is
(a) 480 kPa (b) 804 kPa (c) 1400 kPa (d) 144 kPa (e) 1100 kPa
Answer (b) 804 kPa
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the
following lines on a blank EES screen.
rho=1025 "kg/m^3"
g=9.81 "m/s2"
z1=30 "m"
z2=110 "m"
DELTAP=rho*g*(z2-z1)/1000 "kPa"
,“Some Wrong Solutions with Common Mistakes:”
W1=rho*g*(z2-z1) "not dividing by 1000"
W2=rho*g*(z1+z2)/1000 "adding depts instead of subtracting"
W3=rho*(z1+z2)/1000 "not using g"
W4=rho*g*(0+z2)/1000 "ignoring z1"
Chap1-2 Pressure Difference in Water (Lake)
Consider an 85-m deep lake. The pressure difference between the top and bottom of the
lake is
(a) 834 kPa (b) 85 kPa (c) 417 kPa (d) 1220 kPa (e) 2430 kPa
Answer (a) 834 kPa
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the
following lines on a blank EES screen.
rho=1000 "kg/m^3"
g=9.81 "m/s2"
z1=0 "m"
z2=85 "m"
DELTAP=rho*g*(z2-z1)/1000 "kPa"
“Some Wrong Solutions with Common Mistakes:”
W1=rho*(z1+z2)/1000 "not using g"
,W2=rho*g*(z2-z1)/2000 "taking half of z"
Chap1-3 Pressure Difference in Air (Mountain)
The atmospheric pressures at the top and the bottom of a mountain are read by a
barometer to be 93.8 and 100.5 kPa. If the average density of air is 1.25 kg/m 3, the
height of the mountain is
(a) 5360 m (b) 683 m (c) 547 m (d) 8200 m (e) 7650 m
Answer (c) 547 m
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the
following lines on a blank EES screen.
rho=1.25 "kg/m^3"
g=9.81 "m/s^2"
P1=93.8 "kPa"
P2=100.5 "kPa"
DELTAP=P2-P1 "kPa"
DELTAP=rho*g*h/1000 "kPa"
“Some Wrong Solutions with Common Mistakes:”
DELTAP=rho*W1/1000 "not using g"
DELTAP=g*W2/1000 "not using rho"
P2=rho*g*W3/1000 "ignoring P1"
P1=rho*g*W4/1000 "ignoring P2"
, Chap1-4 Oil Manometer (Duct)
The pressure drop in a duct is to be measured by a differential oil manometer. If the
differential height between the two fluid columns is 3.2 cm and the density of oil is 860
kg/m3, the pressure drop in the duct is
(a) 28 Pa (b) 135 Pa (c) 482 Pa (d) 270 Pa (e) 760 Pa
Answer (d) 270 kPa
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the
following lines on a blank EES screen.
rho=860 "kg/m^3"
g=9.81 "m/s2"
h=0.032 "m"
DELTAP=rho*g*h "Pa"
“Some Wrong Solutions with Common Mistakes:”
W1=rho*h "not using g"
W2=rho*g*h/2 "taking half of z"